You are in the lab and are given two rods set up so that the top rod is directly above the bottom one. The two straight rods 50-cm long are separated by a distance of 1.5-mm each with 15 A of current passing through them. Assume that the bottom rod is fixed and that the top rod is somehow free to move vertically but not horizontally. How much must the top rod weigh so that the system is at equilibrium

Answers

Answer 1

Answer:

Explanation:

The magnetic force due to lower rod must be equal to weight of upper rod for equilibrium .

magnetic field due to lower rod on upper rod

= ( μ₀ / 4π ) x(2i / r ) , i is current , r is distance between rod

= 10⁻⁷ x 2 x 15 / 1.5 x 10⁻³

= 20 x 10⁻⁴ T

force on the upper rod

= B i L , B is magnetic field , i is current in second rod and L is its length

= 20 x 10⁻⁴ x 15 x .50

= 150 x 10⁻⁴ N

= .015 N

This force can balance a wire having weight equal to .015 N .

= .00153 kg

= 1.53 g .

wire should weigh 1.53 g .


Related Questions

interdependence between plants and animals

Answers

Answer:

plants depend on animals for CO2 (to use during photosynthesis) while animals depend on plants for food (consumation)

Explanation:

Answer:

plants depend on animals for CO2 (to use during photosynthesis) while animals depend on plants for food (consumation)

Explanation:

Submarine (sub A) travels through water at a speed of 8.00 m/s, emitting a sonar wave at a frequency of 1 400 Hz. The speed of sound in the water is 1 533 m/s. A second submarine (sub B) is located such that both submarines are traveling directly toward each other. The second submarine is moving at 9.00 m/s. What frequency is detected by an observer riding on sub B as the subs approach each other?

Answers

Answer:

1385 hz

Explanation:

Yes if it’s right pls mark the brainliest

We find that 2N current loops are coplanar and coaxial. The first has radius a and current I. The second has radius 2a and current 2I, and the pattern is repeated up to the Nth, which has radius Na and current NI. The current in the loops alternates in direction from loop to loop as seen from above. Thus the current in the first loop is counterclockwise, in the next clockwise, up to the last loop where it is again clockwise. The magnitude of the magnetic field at the center of the loops is:_________


a. 0

b. µoNI/2a

c. µoI/Na

d. µoNI/a

e. µI/2Na

Answers

Final answer:

The magnitude of the magnetic field at the center of the loops can be determined using Ampere's law. It is given by µ₀NI/(2a), where µ₀ is the permeability of free space, N is the number of turns, I is the current, and a is the radius of the loop.

Explanation:

The magnitude of the magnetic field at the center of the loops can be determined using Ampere's law. According to the law, the magnetic field inside a current-carrying wire of radius a is given by µ₀NI/(2a), where µ₀ is the permeability of free space, N is the number of turns, I is the current, and a is the radius of the loop.

In this case, the radius of the first loop is a and the current is I. Since the pattern is repeated up to the Nth loop, the radius of the Nth loop is Na and the current is NI. Therefore, the magnitude of the magnetic field at the center of the loops is µ₀NI/(2a).

Final answer:

The net magnetic field at the center of a series of alternating coaxial current loops is zero due to the cancellation of the magnetic fields of each pair of loops with opposite currents.

Explanation:

The question refers to a configuration of coaxial current loops whose radii and currents increase with each subsequent loop, and where the direction of current alternates with each loop. To find the magnetic field at the center due to each loop, we use the formula for the magnetic field (B) at the center of a current loop: B = (µ0I)/(2r), where µ0 is the permeability of free space, I is the current, and r is the radius of the loop.

For each pair of consecutive loops, the magnetic fields will be in opposite directions due to the alternating current direction. Thus, the magnetic fields generated by each pair of loops (except for the last one if N is odd) will cancel each other out. If N is even, all fields cancel out and the net magnetic field is zero; if N is odd, the net magnetic field is only due to the largest loop. In both cases, the net magnetic field at the center will be zero.

Therefore, the correct answer is (a) 0.

Sandy and Chris stand on the edge of a cliff and throw identical mass rocks at the same speed. Sandy throw her rock horizontally while Chris throws his upward at an angle of 45° to the horizontal. (i) Which one has a higher speed : Sandy’s thorwn rock at the launch or Chris’s rock when it reaches its highest point in its flight? (ii) Which one (sandy’s rock or Chris’s rock has the higher speed right before hitting the ground? : Explain carefully your answer.

Answers

Final answer:

(i) Sandy's rock has a higher speed at the launch than Chris's rock at its peak. (ii) Both rocks will have the same speed just before hitting the ground due to the conservation of energy.

Explanation:

The subject of this question is physics, specifically pertaining to projectile motion and velocity. To answer this, we need to use the principles of the conservation of energy and the behaviour of objects in projectile motion.

(i) At the point of release, both Sandy's and Chris's rocks have the same initial speed since they were thrown with the same force. However, when Chris's rock reaches the highest point in its flight, its vertical velocity component becomes zero (it stops moving upwards). Despite this, it will retain a horizontal component of velocity, which is equivalent to the initial horizontal speed of the rock thrown by Sandy. Therefore, Sandy's rock will have a higher speed at the point of launch compared to Chris’s rock at its peak.

(ii) Just before hitting the ground, both rocks would have the same speed. This is because the final velocity of a freely falling object is independent of the direction of launch, assuming the object falls to the same height it was thrown from. Conservation of energy ensures that each rock will have the same speed it started with. This remains true irrespective of air resistance.

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(i) Sandy's thrown rock at the launch has a higher speed than Chris's rock when it reaches its highest point in its flight.

(ii) Sandy's rock has the higher speed right before hitting the ground.

(i) At the launch, both Sandy and Chris throw their rocks with the same initial speed [tex]\( v_0 \)[/tex]. Sandy throws her rock horizontally, so its initial vertical speed is 0, and its initial horizontal speed is [tex]\( v_0 \)[/tex]. Chris throws his rock at a 45° angle to the horizontal, which means the initial speed is divided equally between the horizontal and vertical components due to the symmetry of the 45° angle. Therefore, both the horizontal and vertical initial speeds of Chris's rock are [tex]\( \frac{v_0}{\sqrt{2}} \)[/tex].

At the highest point of Chris's rock's trajectory, the vertical component of its velocity becomes 0 because it momentarily stops ascending before starting to fall back down. However, it still has a horizontal component of velocity equal to [tex]\( \frac{v_0}{\sqrt{2}} \)[/tex]. Since [tex]\( \frac{v_0}{\sqrt{2}} \)[/tex] is less than [tex]\( v_0 \)[/tex], Sandy's rock has a higher speed at the launch than Chris's rock at its highest point.

(ii) When both rocks are about to hit the ground, they will both have a vertical component of velocity due to the acceleration of gravity. For Chris's rock, the horizontal component of velocity remains constant throughout the flight (assuming no air resistance), which is [tex]\( \frac{v_0}{\sqrt{2}} \)[/tex]. The vertical component of velocity for Chris's rock will be equal in magnitude to the horizontal component at the moment of release but will have changed direction due to gravity, resulting in a net speed of [tex]\( v_0 \)[/tex] it hits the ground.

Sandy's rock, on the other hand, has been accelerating downwards under gravity while maintaining its initial horizontal speed. Just before hitting the ground, the vertical component of Sandy's rock's velocity will be greater than [tex]\( \frac{v_0}{\sqrt{2}} \)[/tex] due to the acceleration of gravity. The total speed of Sandy's rock just before impact is calculated by combining the horizontal and vertical components using the Pythagorean theorem. Since the horizontal component is [tex]\( v_0 \)[/tex] and the vertical component is greater than [tex]\( \frac{v_0}{\sqrt{2}} \)[/tex], the total speed of Sandy's rock will be greater than [tex]\( v_0 \)[/tex].

Therefore, right before hitting the ground, Sandy's rock has a higher speed than Chris's rock, which has a speed of [tex]\( v_0 \)[/tex]. Sandy's rock's speed is greater because it has both the initial horizontal speed [tex]\( v_0 \)[/tex] and an additional vertical speed component due to the acceleration of gravity.

Where does sheering often occur

Answers

Often occurs most in the crust

A fan is designed to last for a certain time before it will have to be replaced (planned obsolescence). The fan only has one speed (at a maximum of 700 rpm), and it reaches the speed in 4.0 s (starting from rest). It takes the fan 11.0 s for the blade to stop once it is turned off. The manufacturer specifies that the fan will operate up to 1 billion rotations. Andre lives in a hot climate, works outside of the home from approximately 8:00 am to 5:00 pm, Monday through Friday, does not own an air conditioner, and can't sleep with the fan running. Estimate how many hot days ????hot Andre will be able to use the fan, rounded to the nearest day. ????hot= days

Answers

Answer:

1355 days

Explanation:

Find the given attachment

Final answer:

Andre will not be able to use the fan at all during his working hours.

Explanation:

To estimate how many hot days Andre will be able to use the fan, we need to calculate the number of rotations the fan will make during his working hours and subtract the number of rotations it takes for the fan to stop once turned off. We know that the fan reaches a speed of 700 rpm in 4.0 s from rest. This means its angular velocity is 700 * 2π/60 = 73.33 rad/s. So, in 8 hours (Andre's working hours), the fan will rotate for 73.33 * 3600 * 8 = 210,624 rotations. Subtracting the rotations it takes for the fan to stop (700 * 11 = 7700 rotations), Andre will be able to use the fan for approximately 210,624 - 7700 = 202,924 rotations. Since the fan is specified to operate up to 1 billion rotations, Andre will be able to use the fan for approximately 202,924 / 1,000,000,000 = 0.0002029 (or about 0.02%) of its lifespan. Rounded to the nearest day, this is 0 days. 

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A 15m long wire is placed horizontally on the surface of a liquid and is gently pulled up with the force of 60N to keep the wire in equilibrium.what is the surface tension of the liquid?​

Answers

Answer:

2 N/m

Explanation:

Given that

The length of a wire, l = 15 m

Force on the wire, F = 60 N

Surface tension on the liquid, S = ?

The formula to solve this problem is given as

F = S * 2 L, where

F = force on the liquid

S = surface tension of the liquid

L= length of the wire

On substituting the values of each, we have

60 = S * (2 * 15)

S = 60 / 30

S = 2 N/m

Thus, we can say then, that the surface tension of the liquid is 2 N/m

What is the function of a power source in a circuit

Answers

Answer:

Electrical energy

Explanation:

A power source, such as a battery, provides electrical energy for the circuit. The power source is a device or system that converts another form of energy to electrical energy.  

The function of a power source in a circuit is Electrical energy.

Electrical energy:

Electrical energy seems to be a sort of energy (Kinetic) that would be generated by transporting or shifting electric charges. This same quantity of electricity relies somewhat on the velocity of the charges – therefore more electrical energy they represent, the quicker they accelerate.

Examples of Electrical energy are:

Lightning.Batteries.Electric eels.

Thus the response above is the appropriate one.

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A uniform meter stick is pivoted at the 50.00 cm mark on the meter stick. A 400.0 gram object is hung at the 20.0 cm mark on the stick and a 320.0 gram object is hung at the 75.0 cm mark. Drawing is approximate. The meter stick is unbalanced. Determine the cm-mark on the meterstick that a 400 gram object needs to be hung to achieve equilibrium. A) 10.0 B) 40.0 C) 60.0 D) 90.0 E) none of the above is within 10% of my answer

Answers

Answer:C

Explanation:

Given

mass [tex]m_1=400\ gm[/tex] is at [tex]x=20\ cm[/tex] mark

mass [tex]m_2=320\ gm[/tex] is at [tex]x=75\ cm[/tex] mark

Scale is Pivoted at [tex]x=50\ cm mark[/tex]

For scale to be in equilibrium net torque must be equal to zero

Taking ACW as positive thus

[tex]T_{net}=0.4\times g\times (0.5-0.2)-0.32\times g\times(0.75-0.50)[/tex]

[tex]T_{net}=0.12g-0.08g=0.04g[/tex]

Therefore a net torque of 0.04 g is required in CW sense which a mass [tex]400\ gm[/tex] can provide at a distance of [tex]x_o[/tex] from pivot

[tex]0.04g=0.4\times g\times x_o [/tex]

[tex]x_o=0.1\ m[/tex]

therefore in meter stick it is at a distance of [tex]x=60\ cm[/tex]

3. Identify each statement as true or false. Current equals the product of voltage and resistance for an ohmic device.The greater the length of a wire, the higher the resistance of the wire.The potential difference produced by a battery varies depending on the circuit in which it is used.The power produced by a battery varies depending on the circuit in which it is used.The thicker an electrical wire, the higher the resistance of the wire. True False

Answers

Answer:

A) Current equals the product of voltage and resistance for an ohmic device. False (current is voltage divided by resistance)

B) The greater the length of a wire, the higher the resistance of the wire. True (Resistivity is directly proportional to lenght and inversely proportional to the cross sectional area)

C) The potential difference produced by a battery varies depending on the circuit in which it is used. False (the potential difference of a battery is constant, the current drawn is what varies with circuit)

D) The thicker an electrical wire, the higher the resistance of the wire. False ( Resistivity is inversely proportional to the cross sectional area of the conductor).

Final answer:

Current is the voltage divided by resistance, not the product. The resistance of a wire increases with its length and decreases with a larger cross-sectional area. Both the voltage and power output from a battery can vary with the circuit.

Explanation:

True or False: Electrical Principles

Let's evaluate each statement provided:

Current equals the product of voltage and resistance for an ohmic device. False. Current equals the voltage divided by the resistance (I = V/R), according to Ohm's Law.

The greater the length of a wire, the higher the resistance of the wire. True. Resistance is directly proportional to the length of the conductor (R = ρL/A).

The potential difference produced by a battery varies depending on the circuit in which it is used. True. The actual voltage across a battery can vary due to internal resistance and the characteristics of the external circuit.

The power produced by a battery varies depending on the circuit in which it is used. True. Power depends on both the current (I) and the voltage (P = IV), which can be affected by the circuit configuration.

The thicker an electrical wire, the higher the resistance of the wire. False. A thicker wire has a larger cross-sectional area, which reduces resistance (R = ρL/A).

Wonder Woman and Superman fly to an altitude of 1610 km, carrying between them a chest full of jewels that they intend to put into orbit around Earth. They want to make this tempting treasure inaccessible to their evil enemies who are trying to gain possession of it, yet keep it available for themselves for future use when they retire and settle down. But perhaps the time to retire is now! They accidentally drop the chest, which leaves their weary hands at rest, and discover that they are no longer capable of catching it as it falls into the Pacific Ocean. At what speed vf does the chest impact the surface of the water? Ignore air resistance (in reality, it would make large difference). The radius and mass of Earth are 6370 km and 5.98×1024 kg, respectively

Answers

Answer:

5026.55 m/s

Explanation:

Gravitation potential of a body in orbit from the center of the earth is given as

Pg = -GM/R

Where G is the gravitational constant 6.67x10^-11 N-m^2kg^-2

M is the mass of the earth = 5.98x10^24 kg

R is the distance from that point to the center of the earth = r + Re

r is the distance above earth surface, Re is the earth's radius.

R = 1610 km + 6370 km = 7980 km

Pg = -(6.67x10^-11 x 5.98x10^24)/7980x10^3

Pg = -49983208.02 J/kg

The negative sign means that the gravitational potential is higher away from earth than it is at the earth's surface (it shows convention).

This indicates the kinetic energy per kilogram that the chest of jewel will fall with to earth.

Gravitation Potential on earth's surface is

Pg = -GM/Re

= -(6.67x10^-11 x 5.98x10^24)/6370x10^3 = -62616326.53 J/kg

Difference in gravity potential = -49983208.02 - (-62616326.53)

= 12633118.51 J/kg

The velocity V of the jewel chest will be

0.5v^2 = 12633118.51

V^2 = 25266237.02

V = 5026.55 m/s

what is the the wavelength if a wave cycles up and down three times per second and the distance between each wave is 1.7m?

Answers

Answer:5.1 meters

Explanation:

wavelength=3 x1.7

Wavelength=5.1 meters

Beats: A policeman in a stationary car measures the speed of approaching cars by means of an ultrasonic device that emits a sound with a frequency of 39.6 kHz. A car is approaching him at a speed of 35.0 m/s. The wave is reflected by the car and interferes with the emitted sound producing beats. What is the frequency of the beats when the speed of sound in air is 343 m/s?

Answers

Answer:

The frequency of the beats is 43.6408 kHz

Explanation:

Given:

f = frequency = 39.6 kHz

vc = speed of the car = 35 m/s

vs = speed of the sound = 343 m/s

Question: What is the frequency of the beats, f' = ?

As the car moves towards the source, the frequency of the beats

[tex]f'=f*(1+\frac{v_{c} }{v_{s} } )=39.6*(1+\frac{35}{343} )=43.6408kHz[/tex]

A large, 36.0 kg bell is hung from a wooden beam so it can swing back and forth with negligible friction. The bell’s center of mass is 0.55 m below the pivot. The bell’s moment of inertia about an axis at the pivot is 36.0 kg ·m2 . The clapper is a small, 2.8 kg mass attached to one end of a slender rod of length L and negligible mass. The other end of the rod is attached to the inside of the bell; the rod can swing freely about the same axis as the bell. What should be the length L of the clapper rod for the bell to ring silently — that is, for the period of oscillation for the bell to equal that of the clapper?

Answers

Answer:

Length of the rod is 0.043m

Explanation:

Mass of the bell (M) = 36kg

Distance of the center of mass from pivot = 0.55m

Moment of inertia (I) = 36.0kgm²

Mass of clipper (m) = 2.8kg

Length of the bell to ring = L

The period of the pendulum with small amplitude of oscillation is

T = 2π √(I / mgd)

Where g = acceleration due to gravity = 9.8 m/s²

T = 2π √(36 / 36*9.8*0.55)

T = √(0.1855)

T = 0.43s

The period of the pendulum is 0.43s

To find the length of the Clapper rod for which the bell ring slightly,

T = 2π√(L / g)

T² = 4π²L / g

T² *g = 4π²L

L = (T² * g) / 4π²

L = (9.8* 0.43²) / 4π²

L = 1.7287 / 39.478

L = 0.043m

The length of the Clapper rod for the bell to ring slightly is 0.043m

In the development of atomic models, it was realized that the atom is mostly empty. Consider a model for the hydrogen atom where its nucleus is a sphere with a radius of roughly 10−15 m, and assume the electron orbits in a circle with a radius of roughly 10−10 m. In order to get a better sense for the emptiness of the atom, choose an object and estimate its width. This object will be your "nucleus". How far away would the "electron" be located away from your "nucleus"? Choose an object and estimate its width. Give an answer between 0.0001 m and 100 m. m Using your estimated width as the diameter of the "nucleus" and calculate the radius of the "electron's" orbit. m

Answers

Answer:

Check the explanation

Explanation:

a) in solving the first question, we will be choosing a spherical ball of radius 1 m, Therefore the width of the object is the diameter = 2m.

b)Given that and according to the question, the radius of the electron's orbit = 1x105 x radius of the object = 1x105 x1 = 1x105 m

A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.80 rad????s. Its total moment of inertia is 1360 kg ???? m2. Four people standing on the ground, each of mass 65 kg, suddenly step onto the edge of the merry-go-round. (a) What is the angular velocity of the merry-go-round now? (b) What if the people were on it initially and then jumped off in a radial direction (relative to the merry-go-round)

Answers

Final answer:

This question is about the principle of the conservation of angular momentum and how changes in moment of inertia affect angular velocity. When people step onto the merry-go-round, its moment of inertia increases causing the angular velocity to decrease, whereas when they jump off, the moment of inertia decreases, thus increasing the angular velocity.

Explanation:

This question falls into the realm of rotational dynamics, specifically dealing with the conservation of angular momentum and the effect of moment of inertia on angular velocity. Let's break it down:

(a) Four people step onto the merry-go-round

When the four people step onto the merry-go-round, they increase its moment of inertia, thus reducing its angular velocity according to the principle of conservation of angular momentum. We can find the new angular velocity using the equation L_initial = L_final, where L is the angular momentum. Hence, we have I_initial * ω_initial = I_final * ω_final. Rewriting for ω_final gives us ω_final = (I_initial * ω_initial / I_final).

(b) People jump off the merry-go-round

On the other hand, when people jump off the merry-go-round, they are reducing its moment of inertia, thus increasing the angular velocity under the conservation of angular momentum. Hence, a similar equation would apply, only that the final angular velocity would be higher than the initial angular velocity because the final moment of inertia is lower due to the people jumping off.

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Find the magnitude of the average force ⟨Fx⟩⟨Fx⟩ in the x direction that the particle exerts on the right-hand wall of the container as it bounces back and forth. Assume that collisions between the wall and particle are elastic and that the position of the container is fixed. Be careful of the sign of your answer. Express the magnitude of the average force in terms of mmm, vxvxv_x, and LxLxL_x.

Answers

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

The attached file has a detailed solution of the given problem.

Newton's second law allows us to find the average force for the impact of a particle against the wall of a container is:

        [tex]F = \frac{m v_x^2}{L_x}[/tex]  

Newton's second law is the change of the momentum with respect to time and the momentum is defined as the product of the mass and the velocity of the body.

           [tex]F = \frac{dp}{dt} \\p = m v[/tex]

where F is the force, p the moment, m the mass, t the time and v is the velocity.

Ask for the average force, therefore we change the differentials for variations.

           [tex]<F> = m \frac{\Delta v}{\Delta t}[/tex]  

They indicate that the velocity in the direction of the wall is vₓ and the mass of the container is much greater than the mass of the particle, they also indicate that the collision is elastic, therefore the speed of the particle before and after the collision is equal , but its address changes.

      Δv = vₓ - (-vₓ)

      Δv = 2 vₓ

The change in velocity occurs during the collision, in the rest of the motion the particle has a constant velocity, using the uniform motion relation.

         [tex]v = \frac{d}{t} \\t = \frac{d}{v}[/tex]  

The particle travels a distance Lₓ from inside the container to the wall and bounces, we can find the total time for the particle where the distance of the entire route is:

          d = 2 Lₓ

          t = [tex]\frac{2L_x}{v_x}[/tex]

Let's substitute in Newton's second law.

     [tex]<F>= \frac{m \ 2v_x}{ \frac{2L_x}{v_x} } \\<F>= \frac{m v_x^2}{L_x}[/tex]

In conclusion, using Newton's second law we can find the average force for the collision of a particle against the wall of a container is:

         [tex]<F> = \frac{m \ v_x^2}{L_x}[/tex]  

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The noise level coming from a pig pen with
199 pigs is 74.3 dB. The intensity of the sound
coming from the pig pen is proportional to the
number of pigs.
Assuming each of the remaining pigs squeal
at their original level after 61 of their compan-
ions have been removed, what is the decibel
level of the remaining pigs?
Answer in units of dB.

Answers

The decibel level of the remaining pigs is 34.02dB.

To solve this problem, we need to understand the relationship between the number of pigs and the intensity of sound. Since the intensity is proportional to the number of pigs, we can use the formula:

[tex]\[ \text{Intensity} = k \times \text{Number of pigs} \][/tex]

where ( k ) is a constant of proportionality.

Given that the noise level from 199 pigs is 74.3 dB, we can set up the equation:

[tex]\[ 74.3 = k \times 199 \][/tex]

First, let's solve for ( k ):

[tex]\[ k = \frac{74.3}{199} \][/tex]

Now, we can use this value of ( k ) to find the intensity when only 138 pigs remain:

[tex]\[ \text{Intensity}_{\text{new}} = k \times 138 \][/tex]

Finally, we can convert this intensity back into decibels using the formula:

[tex]\[ \text{Noise level (dB)} = 10 \log_{10}(\text{Intensity}) \][/tex]

Now, let's calculate:

[tex]\[ k = \frac{74.3}{199} \][/tex]

[tex]\[ k \approx 0.3726 \][/tex]

[tex]\[ \text{Intensity}_{\text{new}} = 0.3726 \times 138 \][/tex]

[tex]\[ \text{Intensity}_{\text{new}} = 51.36 \][/tex]

[tex]\[ \text{Noise level (dB)} = 10 \log_{10}(51.36) \][/tex]

[tex]\[ \text{Noise level (dB)} = 34.02 \][/tex]

So, the decibel level of the remaining pigs is 34.02dB.

An ammeter is connected in series with a resistor of unknown resistance R and a DC power supply of known emf e. A student uses the ammeter to measure the current I and calculates the power P dissipated by the resistor using I and e. The student later discovers that the ammeter is not an ideal ammeter. Which of the following correctly compares the actual power dissipated by the resistor to the value of power P calculated by the student and explains why? (A) The actual power dissipated by the resistor is greater than P because the ammeter had some resistance (B) The actual power dissipated by the resistor is less than P because the ammeter had some resistance. (C) The actual power dissipated by the resistor is greater than P because the ammeter should have been connected in parallel to the resistor. (D) The actual power dissipated by the resistor is less than P because the ammeter should have been connected in parallel to the resistor. (E) The actual power dissipated by the resistor is equal to P because the ammeter was connected properly in series.

Answers

Answer:

B:The actual power dissipated by the resistor is less than P because the ammeter had some resistance.

Explanation:

Here,power has been calculated using current I and total EMF \ε . So,P=EMF*current= ε I will represent total power dissipated in resistor and ammeter.

Now, this total power P has been dissipated in both resistor and ammeter. So, power dissipated in resistor must be less than P as some power is also dissipated in ammeter because it has non-zero resistance.

So, the answer is B:The actual power dissipated by the resistor is less than P because the ammeter had some resistance.

Note that option A,C and E are ruled out as they state power dissipated by resistor is greater than or equal to P which is false.

Also,option D is ruled out as ammeter is connected in series.

g The steam above a freshly made cup of instant coffee is really water vapor droplets condensing after evaporating from the hot coffee. What is the final temperature (in °C) of 240 g of hot coffee initially at 80.0°C if 2.50 g evaporates from it? The coffee is in a Styrofoam cup, and so other methods of heat transfer can be neglected. Assume that coffee has the same physical properties as water.

Answers

Answer:

the final temperature = 74.33°C

Explanation:

Using the expression Q = mcΔT for the heat transfer and the change in temperature .

Here ;

Q = heat transfer

m = mass of substance

c = specific heat

ΔT = the change in temperature

The heat Q required to change the phase of a sample mass  m is:

Q = m[tex]L_v[/tex]

where;

[tex]L_v[/tex]  is the latent heat of vaporization.

From the question ;

Let M represent the mass of the coffee that remains after evaporation is:

ΔT = [tex]\frac{mL_v}{MC}[/tex]

where;

m = 2.50 g

M = (240 - 2.50) g  = 237.5 g

[tex]L_v[/tex]  = 539 kcal/kg

c = 1.00kcal/kg. °C

ΔT = [tex]\frac{2.50*539 \ kcal /kg}{237.5 g *1.00 \ kcal/kg . ^0C}[/tex]

ΔT = 5.67°C

The final temperature of the coffee is:

[tex]T_f = T_i -[/tex] ΔT

where ;

[tex]T_I[/tex] = initial temperature = 80 °C

[tex]T_f[/tex] = (80 - 5.67)°C

[tex]T_f[/tex] =  74.33°C

Thus; the final temperature = 74.33°C

Answer:

Final temperature of the hot coffee, [tex]\theta_{f} = 85.67^0 C[/tex]

Explanation:

Mass of coffee remaining after evaporation of 2.50 g, M = 240 - 2.50

M = 237.5 g

Mass of evaporated coffee, m = 2.5 g

Initial temperature of hot coffee, [tex]\theta_{i} = 80^{0} C[/tex]

Initial temperature of hot coffee, [tex]\theta_{f} =[/tex] ?

Let the specific heat capacity of the coffee, c = 1 kcal/kg

Latent heat of vaporization of coffee, [tex]L_{v} = 539 kcal/kg[/tex]

The heat energy due to temperature change:

[tex]Q = Mc \triangle \theta[/tex]

[tex]Q = 237.5 * 1 * \triangle \theta[/tex]...........(1)

The heat energy due to change in state

[tex]Q = mL_{v}[/tex]

Q = 2.5 * 539

Q = 1347.5..........(2)

Equating (1) and (2)

[tex]1347.5 = 237.5 \triangle \theta\\\triangle \theta = 1347.5/237.5\\\triangle \theta =5.67^{0} C[/tex]

[tex]\triangle \theta = \theta_{f} - \theta_{i} \\5.67 = \theta_{f} - 80\\\theta_{f} = 80 + 5.67\\\theta_{f} = 85.67^0 C[/tex]

Identify three pollutants released into the air when fossil fuels are burned.

Answers

Answer:

Carbon Monoxide (CO), Carbon Dioxide (CO2), and Sulfur Dioxide (SO2).

Answer:Carbon dioxide (CO2), Carbon monoxide (CO), Methane (CH4)

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The ball rolling along a straight and level path. The ball is rolling at a constant
speed of 2.0 m/s. The mass of the ball is 3.0 kg. What is the kinetic energy of the
ball? (KE=0.5*m*v)

Answers

Answer:6 joules

Explanation:

Mass(m)=3kg

Velocity(v)=2m/s

Kinetic energy=0.5 x m x v^2

Kinetic energy=0.5 x 3 x 2^2

Kinetic energy=0.5 x 3 x 2 x 2

Kinetic energy=6

A narrow beam of light from a laser travels through air (n = 1.00) and strikes the surface of the water (n = 1.33) in a lake at point A. The angle of incidence is 70 degrees. The depth of the lake is 4.3 m. On the flat lake-bottom is point B, directly below point A. (a) If refraction did not occur, how far away from point B would the laser beam strike the lakebottom? (b) Considering refraction, how far away from point B would the laser beam strike the lake-bottom?

Answers

Answer:

A) d = 11.8m

B) d = 4.293 m

Explanation:

A) We are told that the angle of incidence;θ_i = 70°.

Now, if refraction doesn't occur, the angle of the light continues to be 70° in the water relative to the normal. Thus;

tan 70° = d/4.3m

Where d is the distance from point B at which the laser beam would strike the lakebottom.

So,d = 4.3*tan70

d = 11.8m

B) Since the light is moving from air (n1=1.00) to water (n2=1.33), we can use Snell's law to find the angle of refraction(θ_r)

So,

n1*sinθ_i = n2*sinθ_r

Thus; sinθ_r = (n1*sinθ_i)/n2

sinθ_r = (1 * sin70)/1.33

sinθ_r = 0.7065

θ_r = sin^(-1)0.7065

θ_r = 44.95°

Thus; xonsidering refraction, distance from point B at which the laser beam strikes the lake-bottom is calculated from;

d = 4.3 tan44.95

d = 4.293 m

Scientific work is currently underway to determine whether weak oscillating magnetic fields can affect human health. For example, one study found that drivers of trains had a higher incidence of blood cancer than other railway workers, possibly due to long exposure to mechanical devices in the train engine cab. Consider a magnetic field of magnitude 1.00 10-3 T, oscillating sinusoidally at 72.5 Hz. If the diameter of a red blood cell is 7.60 µm, determine the maximum emf that can be generated around the perimeter of a cell in this field.

Answers

Given Information:

Magnetic field = B = 1×10⁻³ T

Frequency = f = 72.5 Hz

Diameter of cell = d = 7.60 µm = 7.60×10⁻⁶ m

Required Information:

Maximum Emf = ?

Answer:

Maximum Emf = 20.66×10⁻¹² volts

Explanation:

The maximum emf generated around the perimeter of a cell in a field is given by

Emf = BAωcos(ωt)

Where A is the area, B is the magnetic field and ω is frequency in rad/sec

For maximum emf cos(ωt) = 1

Emf = BAω

Area is given by

A = πr²

A = π(d/2)²

A = π(7.60×10⁻⁶/2)²

A = 45.36×10⁻¹² m²

We know that,

ω = 2πf

ω = 2π(72.5)

ω = 455.53 rad/sec

Finally, the emf is,

Emf = BAω

Emf = 1×10⁻³*45.36×10⁻¹²*455.53

Emf = 20.66×10⁻¹² volts

Therefore, the maximum emf generated around the perimeter of the cell is 20.66×10⁻¹² volts

Which side of the electromagnetic spectrum has shorter wavelengths?
Which side of the electromagnetic spectrum has higher energy?)
Which side of the electromagnetic spectrum has higher frequencies?

Answers

Gamma rays have the smallest wavelengths and the most energy of any wave in the electromagnetic spectrum. They are produced by the hottest and most energetic objects in the universe, such as neutron stars and pulsars, supernova explosions, and regions around black holes.

What is wavelength ?

"Distance between corresponding points of two consecutive waves." “Corresponding points” refers to two points or particles in the same phase—i.e. points that have completed identical fractions of their periodic motion.

What is frequency?

"Tthe number of waves that pass a fixed point in unit time. "It also describes the number of cycles or vibrations undergone during one unit of time by a body in periodic motion.

Know more about gamma rays here

https://brainly.com/question/23281551

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Answer:

1.blue

2.blue

3.blue

Explanation:

I took the quiz

You are suffering through a court case arising from a recent car accident that you were in. You were traveling West, alone in your car, which has a mass of 1000 kg, through an intersection when another driver in a small car (mass 850 kg) traveling South crashed into your passenger side at the center of the intersection. The two cars became stuck together and skidded off the road. For this problem, choose the positive x-axis to point to the East (right) and the positive y-axis to point North (up). You recall that you had just looked at the speedometer before the accident and were traveling at 20 m/s. What was your initial momentum vector?

Answers

Answer:

p₀ = (- 20,000 i - 17,000 j) kg m / s

vₓ = -10.81 m / s ,   v_{y} = - 9,189 m / s  

Explanation:

This exercise can be solved with the conservation of the moment.

The system is formed by the two cars, so the forces during the crash are internal and the moment is conserved

initial moment. Just before the crash

      p₀ = M v₁ + m v₂

where the masses M = 1000 kg with a velocity of v₁ = - 20 i m / s for traveling west, the second vehicle has a mass m = 850 kg and a velocity of v₂ = - 20 i m / s, we substitute

     p₀ = - 1000 20 i - 850 20 j

     p₀ = (- 20,000 i - 17,000 j) kg m / s

final moment. Right after the crash

how the cars fit together

      [tex]p_{f}[/tex] = (M + m) [tex]v_{f}[/tex]

the final speed is shaped

      v_{f} = vₓ i + v_{y} j

  we substitute

     p_{f} = (M + m) (vₓ i + v_{y} j)

     p_{f} = (1000 + 850) (vₓ i + [tex]v_{y}[/tex]j)

   

     p₀ = p_{f}

      (- 20000 i - 17000 j) = 1850 (vₓ i + v_{y} j)

we solve for each component

    vₓ = -20000/1850

    vₓ = -10.81 m / s

    v_{y} = -17000/1850

    v_{y} = - 9,189 m / s



In order for a current to be maintained through a conductor a difference in ________________________ must be maintained from one end of the conductor to the other end.

Answers

Answer:

Potential

Explanation:

In order for a current to be maintained through a conductor a difference in potential must be maintained from one end of the conductor to the other end. Due to potential difference and emf is induced in it. As a result current starts to flow.

A single slit forms a diffraction pattern, with the first minimum at an angle of 40 degrees from central maximum. Monochromatic light of 410 nm wavelength is used. The same slit, illuminated by a different monochromatic light source, produces a diffraction pattern with the second minimum at a 60 degree angle from the central maximum. The wavelength of this light, in nm, is closest to:

Answers

Answer:

276.19nm

Explanation:

To find the other wavelength you use the following condition for the diffraction of both wavelengths:

[tex]m_1\lambda_1=asin\theta_1\\\\m_2\lambda_2=asin\theta_2\\\\[/tex]    ( 1 )

λ1=410nm

m=1 for wavelength 1

m=2 for wavelength 2

a: width of the slit

θ1: angle of the first minimum

θ2: angle of the second minimum

you divide both equations and you obtain:

[tex]\frac{m_1\lambda_1}{m_2\lambda_2}=\frac{sin\theta_1}{sin\theta_2}\\\\\lambda_2=\frac{sin\theta_2}{sin\theta_1}\frac{m_1\lambda_1}{m_2}\\\\\lambda_2=\frac{sin60\°}{sin40\°} \frac{(1)(410nm)}{2}=276.19nm[/tex]

hence, the wavelength of the second monochromatic wave is 276.19nm

Answer:

The wavelength of the  second monochromatic light is  [tex]\lambda = 277 nm[/tex]

Explanation:

From the question we are told that

    The angle of the first minimum is [tex]\theta_1 = 40^o[/tex]

   The wavelength of the first  monochromatic light is  [tex]\lambda_1 = 410 \ nm[/tex]

    The angle of the second minima is  [tex]\theta_2 = 60^o[/tex]

     

For the first minima the distance of separation of diffraction patterns is mathematically represented as

       [tex]a = \frac{\lambda_1 }{sin \theta_1}[/tex]

 Substituting values

       [tex]a = \frac{410 *10^{-9}}{sin (40) }[/tex]

       [tex]a = 638 nm[/tex]

The distance between two successive diffraction is constant for the same slit

  Thus the wavelength of the second light is

             [tex]\lambda = \frac{a * sin (60)}{2}[/tex]

 Substituting value

              [tex]\lambda = \frac{638 * sin (60)}{2}[/tex]

              [tex]\lambda = 277 nm[/tex]

         

A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 3.59 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2190 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) in the positive z direction and has a magnitude of 3.91 V/m, (b) in the negative z direction and has a magnitude of 3.91 V/m, and (c) in the positive x direction and has a magnitude of 3.91 V/m?

Answers

Answer:

a) 1.88*10^-18 N

b) 6.32*10^-19 N

c) 1.9*10^-18 N

Explanation:

The total force over the electron is given by:

[tex]\vec{F}=q\vec{E}+q\vec{v}\ X\ \vec{B}[/tex]

the first term is the electric force and the second one is the magnetic force.

You have that the velocity of the electron and the magnetic field are:

[tex]\vec{v}=2190\frac{m}{s}\ \hat{j}\\\\\vec{B}=-5.39*10^{-3}T\ \hat{i}[/tex]

by using the relation  j X (-i) =- j X i = -(-k) = k, you obtain:

[tex]\vec{v} \ X\ \vec{B}=(2190m/s)(3.59*10^{-3}T)\hat{k}=7.862\ T m/s[/tex]

a) For an electric field of 3.91V/m in +z direction:

[tex]\vec{F}=q[\vec{E}+\vec{v}\ X\ \vec{B}]=(1.6*10^{-19})[3.91\hat{k}+7.862\ \hat{k}]N\\\\\vec{F}=1.88*10^{-18}N\hat {k}\\\\F=1.88*10^{-18}N[/tex]

b) E=3.91V/m in -z direction:

[tex]\vec{F}=(1.6*10^{-19})[-3.91\hat{k}+7.862\ \hat{k}]N\\\\\vec{F}=6.32*10^{-19}N\hat {k}\\\\F=6.32*10^{-19}N[/tex]

c) E=3.91 V/m in +x direction:

[tex]\vec{F}=(1.6*10^{-19})[3.91\hat{i}+7.862\ \hat{k}]N\\\\\vec{F}=[6.25*10^{-19}\ \hat{i}+1.25*10^{-18}\ \hat{k} ]N\\\\F=\sqrt{(6.25*10^{-19})^2+(1.25*10^{-18})^2}N=1.9*10^{-18}\ N[/tex]

Using this formula , What is the weight, on Earth, of an object with a mass
of 72.4kg?*
716.52N
659.52N
739.52N
709.52N

Answers

Answer:

W = 709.52 N

Explanation:

It is given that,

The mass of an object on Earth is 72.4 kg.

We need to find the weight of the object on Earth.

The formula of the weight of an object is given by :

W = mg

g is acceleration due to gravity on Earth, g = 9.8 m/s²

So,

[tex]W = 72.4\times 9.8\\\\W=709.52\ N[/tex]

So, the weight of the object is 709.52 N.

Answer:

So, the weight of the object is 709.52 N.

Explanation:

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